Derivatives, rates of change (triangle and angle)

[physics604]

1. Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle
between the sides of fixed length is \pi/3.

Homework Equations

$$A=\frac{xysinθ}{2}$$

The Attempt at a Solution

Given:
$$\frac{dθ}{dt}=0.06$$ $$θ=\frac{\pi}{3}$$ $$x=4$$ $$y=5$$
Find: $$\frac{dA}{dt}$$

$$2A=xysinθ$$ $$2lnA=lnxysinθ$$ $$2\frac{1}{A}\frac{dA}{dt}=lnx+lny+lnsinθ$$

The problem now is that everything on the right cancels out, because x and y are constant, and the derivative sinθ is 0.

Am I doing this correctly? Where can I go from here?

 

[Simon Bridge]

You already have $$A(t)=10\sin\!\big(\theta (t)\big)$$ since xy/2=10.
What's wrong with differentiating both sides wrt t?

Notes:

$\ln(2A)\neq 2\ln(A)$

$\frac{d}{dt}\sin\theta \neq 0$ because $\theta$ is a function of time.

 

[physics604]

Okay, but I still get

$$A=10sinθ$$ $$\frac{dA}{dt}=10\frac{dsinθ}{dt}$$

 

[Simon Bridge]

Hint: chain rule

Aside: in LaTeX, you format special functions by putting a backsash in front of the abbreviation
i.e \ln(x) becomes $\ln(x)$ and \sin\theta becomes $\sin\theta$ ... cool huh?

 

[physics604]

$$\frac{dsinθ}{dt}=\frac{d}{dx}\frac{sinθ}{t}?$$

Is this correct? I really have no idea what to do with it.

 

[Simon Bridge]

No - theta is a function of time.

The chain rule says $$\frac{d}{dt}f(g(t)) = \frac{df}{dg}\frac{dg}{dt}$$

 

[Ray Vickson]

$$\frac{dsinθ}{dt}=\frac{d}{dx}\frac{sinθ}{t}?$$

Is this correct? I really have no idea what to do with it.

Do you know the formula for the derivative of the sine function; that is, can you say what is
\frac{d}{dw} \sin(w)\:?
If not, you need to go back to the very beginning of your calculus notes, or maybe you are taking material that is beyond your background, If you DO know it, just use it in the chain rule.

 

[physics604]

cosw

 

[physics604]

$$cosθ(t)×\frac{dθ}{dt}?$$

 

[Simon Bridge]

Now you've got it :)